# Integration of Rational Functions by partial fractions

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• Feb 21st 2008, 05:09 PM
galactus
With trig sub it whittlles down nicely.

Let $\displaystyle x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}$

Make the subs and you get:

$\displaystyle \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}$
• Feb 21st 2008, 05:12 PM
FalconPUNCH!
Quote:

Originally Posted by galactus
With trig sub it whittlles down nicely.

Let $\displaystyle x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}$

Make the subs and you get:

$\displaystyle \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}$

yeah I did it that way, but I got stuck on the limits of integration (Worried) I don't know what $\displaystyle 1 = sec({\theta})$ is. Is there a quick way to figure out what they are?
• Feb 21st 2008, 05:14 PM
topsquark
Quote:

Originally Posted by FalconPUNCH!
yeah I did it that way, but I got stuck on the limits of integration (Worried) I don't know what $\displaystyle 1 = sec({\theta})$ is. Is there a quick way to figure out what they are?

$\displaystyle sec(\theta) = 1$

$\displaystyle \frac{1}{cos(\theta)} = 1$

$\displaystyle cos(\theta) = 1$

$\displaystyle \theta = 0$

-Dan
• Feb 21st 2008, 05:15 PM
mr fantastic
Quote:

Originally Posted by FalconPUNCH!
I get $\displaystyle \frac{40}{3}$

hmmm...seem to be getting the limits of integration wrong which is happening in my homework a lot.

for this one $\displaystyle \int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x}dx$ is it best to use u substitution from the start?

Alternatively, make the substitution $\displaystyle u^2 = x^2 - 1$.

Then the integral becomes

$\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$.

Details can be given if needed .....
• Feb 21st 2008, 05:25 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Alternatively, make the substitution $\displaystyle u^2 = x^2 - 1$.

Then the integral becomes

$\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$.

Details can be given if needed .....

$\displaystyle u^2 = x^2 - 1 \Rightarrow 2u \frac{du}{dx} = 2x \Rightarrow dx = \frac{u}{x} \, du$

$\displaystyle x = 1 \Rightarrow u^2 = 0 \Rightarrow u = 0$. $\displaystyle x = 2 \Rightarrow u^2 = 3 \Rightarrow u = \sqrt{3}$ where the positive root is used.

Then:

$\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u}{x} \frac{u}{x} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{x^2} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du$

$\displaystyle = \int_{0}^{\sqrt{3}} \frac{(u^2 + 1) - 1}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$,

which integrates easily to give $\displaystyle I = \sqrt{3} - \frac{\pi}{3}$.
• Feb 21st 2008, 05:42 PM
FalconPUNCH!
Thank you guys for helping me out. I'm going to use this thread as a reference so I won't forget how to do these. (Handshake)
• Feb 21st 2008, 05:50 PM
mr fantastic
Quote:

Originally Posted by FalconPUNCH!
Thank you guys for helping me out. I'm going to use this thread as a reference so I won't forget how to do these. (Handshake)

Good thinking, 99 (Rofl)
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